What is the relationship between pressure and flow rate?

[HystereeSis]

Homework Statement

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Relevant Equations

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I performed an experiment where I let water flow out of a measuring cylinder through a hole at the base of the measuring cylinder. I let the water run out for 30s, recording the volume after. I did this for different heights of water in the measuring cylinder, and kept the pressure of water at the top of the hole constant by continually adding water to the measuring cylinder (in order to keep the height of water the same). I calculated the flow rate using vol / time and found pressure using P = rho g h. I plotted a graph of flow rate by pressure and got a straight line through the origin, suggesting that flow rate and pressure are directly proportional. Does anyone know a mathematical relationship in terms of fluid dynamics to explain this? I was thinking perhaps Bernoulli's equation but I'm having trouble understanding that as (among other misunderstandings) I have calculated flow rate, not velocity of the water.

I would be so grateful for any help.

 

[gmax137]

Yes, Bernoulli is the way, at least for a simple estimate of the flow rate. More specifically, look up Torricelli's Law to get started.

As to velocity, if you can measure the area of the hole, you can then relate flow rate to velocity. If it is a nice round hole, measure the diameter and calculate the area.

 

[HystereeSis]

Yes, Bernoulli is the way, at least for a simple estimate of the flow rate. More specifically, look up Torricelli's Law to get started.

As to velocity, if you can measure the area of the hole, you can then relate flow rate to velocity. If it is a nice round hole, measure the diameter and calculate the area.

Torricelli's Law: discharge velocity = √(2gh)
I'm supremely confused as to how this relates to Bernoulli, unfortunately :/
Bernoulli's equation: P + 1/2 rho v^2 + rho g h = constant
I read somewhere that P is static pressure and rho g h is hydrostatic pressure.
In my calculations (x axis of graph), I used rho g h - i.e. density of water at room temp, g, height of water in the vertical measuring cylinder. Is this sound?
I just have no clue how to use / rearrange Bernoulli's equation to show pressure as directly proportional to flow rate.

 

[gmax137]

Read the wiki page on torricelli. It seems pretty good.

The idea with Bernoulli is, you have P + 1/2 rho v^2 + rho g h = constant. So it has the same value at the top surface (in your measuring cylinder) and at the exit of the hole. So if you call the top surface "1" and the hole "2" you can write an equation

$$P_1 + {1/2}~ \rho ~ {v_1}^2 + \rho ~g ~h_1 = P_2 + {1/2}~ \rho ~{v_2}^2 + \rho ~ g ~ h_2$$

then solve for v2. The Wiki page shows how it is done.

EDIT: Latex fixes

 

[HystereeSis]

Read the wiki page on torricelli. It seems pretty good.

The idea with Bernoulli is, you have P + 1/2 rho v^2 + rho g h = constant. So it has the same value at the top surface (in your measuring cylinder) and at the exit of the hole. So if you call the top surface "1" and the hole "2" you can write an equation

$$P_1 + {1/2}~ \rho ~ {v_1}^2 + \rho ~g ~h_1 = P_2 + {1/2}~ \rho ~{v_2}^2 + \rho ~ g ~ h_2$$

then solve for v2. The Wiki page shows how it is done.

EDIT: Latex fixes

Would you mind replying with the link to said web page?

 

[gmax137]

Sure:
https://en.wikipedia.org/wiki/Torricelli's_law

Hope that helps!

 

[gmax137]

Also note, it does not show pressure proportional to flow rate. There's a square root in there, you should get

$flow~rate~ \sim ~ \sqrt h$

 

[Blanchdog]

Alright so you have taken steps to maintain constant pressure and thus constant flow rate at each height of water. If I understand your question correctly, you're looking for a linear relationship between pressure and flow rate. Assuming a nice minimally turbulent flow, this should imply a similar relationship to exit velocity. Let's use Bernoulli's equation and see if it turns out to be a linear relationship.

$P + \frac{1}{2}\rho {v_1}^2 + \rho g h_1 = P + \frac{1}{2}\rho {v_2}^2 + \rho g h_2$

We will assume atmospheric pressure is the same at the top of the tank and at the exit hole. We will make a simplifying approximation that the surface area of the tank is large and thus the downward velocity small enough to call zero. We define the height of the drain hole to be zero, the height of the water column to be h, and the exit velocity to be v. Bernoulli's equation then becomes

$\frac{1}{2}\rho {v}^2 = \rho g h \rightarrow \frac{1}{2} {v}^2 = g h \rightarrow v = \sqrt{2 g h}$

This final expression is commonly known as Toriccelli's Law, and shows that the relationship between exit velocity and water column height is nonlinear, and by extension, the relationship between flow rate and pressure due to water column height is nonlinear. If you used a much taller column of water you would find that a linear prediction based on your data would overestimate the flow rate.

 

[HystereeSis]

Thank you both so so so so so much.